Question

How do you integrate `int 1/sqrt(9x^2-36x+37)` using trig substitutions?

Answer 1

By substituting `x=u/3+2` the square root becomes `sqrt(u^2+1)` which is standard form or can be further substituted with `u=sinhv` and then use `cosh^2 v - sinh^2v=1`, `d/dx(sinh v)=cosh(v)` etc.